Ch13Word - Chapter 13 CHAPTER 13 Fluids 1 2 3 4 When we use...

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Chapter  13 CHAPTER 13 – Fluids 1. When we use the density of granite, we have m  =  ρ V  = (2.7 × 10 3  kg/m 3 )(1 × 10 8   m 3 ) =        3 × 10 11  kg . 2. When we use the density of air, we have m  =  ρ V  =  ρ LWH  = (1.29 kg/m 3 )(4.8 m)(3.8 m)(2.8 m) =        66 kg . 3. When we use the density of gold, we have m  =  ρ V  =  ρ LWH  = (19.3 × 10 3  kg/m 3 )(0.60 m)(0.25 m)(0.15 m) =        4.3 × 10 2  kg        (˜ 950 lb!). 4. If we assume  a mass of 65 kg with the density of water, we have m  =  ρ V ; 65 kg = (1.0 × 10 3  kg/m 3 ) V , which gives  V  =        6.5 × 10 –2  m 3         (˜ 65 L). 5. From the masses we have m water  = 98.44 g – 35.00 g = 63.44 g; m fluid  = 88.78 g – 35.00 g = 53.78 g. Because the water and  the fluid occupy the same volume, we have SG fluid  =  ρ fluid / ρ water  =  m fluid / m water  = (53.78 g)/(63.44 g) =        0.8477 . 6. The definition of the specific gravity of the mixture is SG mixture  =  ρ mixture / ρ water . The density of the mixture is  ρ mixture  =  m mixture / V  = ( SG antifreeze ρ water V antifreeze  +  SG water ρ water V water )/ V , so we get SG mixture   = ( SG antifreeze V antifreeze  +  SG water V water )/ V = [(0.80)(5.0 L) + (1.0)(4.0 L)]/(9.0 L) =        0.89 . 7. ( a ) The normal force on the four legs must equal the weight.  The pressure of the reaction to the normal  force, which is exerted  on the floor, is P  =  F N / A  =  mg / A  = (60 kg)(9.80 m/s 2 )/4(0.05 × 10 –4  m 2 ) =        3 × 10 7  N/m 2 . ( b ) For the elephant  standing  on one foot, we have P  =  F N / A  =  mg / A  = (1500 kg)(9.80 m/s 2 )/(800 × 10 –4  m 2 ) =        2 × 10 5  N/m 2 . Note that this is a factor of ˜ 100 ×  less than  that of the loudspeaker! 8. ( a ) The force of the air on the table top is F  =  PA  = (1.013 × 10 5  N/m 2 )(1.6 m)(2.9 m) =        4.7 × 10 5  N (down) . ( b ) Because the pressure  is the same on the underside  of the table, the upward  force has the same  magnitude:        4.7 × 10 5  N .        This is why  the table does not move! 9. The pressure  difference on the lungs is the pressure  change from the depth  of water:  ? P  =  ρ ? h ; (80 mm-Hg)(133 N/m 2     mm-Hg) = (1.00 × 10 3  kg/m 3 )(9.80 m/s 2 )? h , which gives ? h  =        1.1 m . 10. There is atmospheric pressure  outside  the tire, so we find the net force from the gauge pressure.  Because the  reaction to the force from the pressure  on the four footprints of the tires supports the automobile, we have  4 PA  =  mg ; 4(240 × 10 3  N/m 2  )(200 × 10 –4  m 2 ) =  m (9.80 m/s 2 ), which gives  m  =       2.0 × 10 3  kg .
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